Diodes - red & orange (light from one or both)

Depending on which substances they're composed, light emitting diodes typically need 1.5 - 3 V in forward bias to give satisfactory light. You have two light emitting diodes, red and orange, you want to couple together in a capsule. You want to make a circuit where the capsule sends out red or orange light, or a mixture of these two. That is so that one or the other or both diodes send out light. Draw ciruits with the diodes and a proper power supply.

2. Relevant equations

N/A. That is I must use Kirchoffs 2. law for the voltages, right?

3. The attempt at a solution

I don't know exactly how to solve this.

The diodes need 1.5 - 3 V voltage. To make the different combinations (diode 1 sends out red light, diode 2 sends out orange light, or they both combine to give a red-orange light) I must vary series and parallell coupling right? Or shall all combinations be possible in the same circuit?

Problem: An LED emits photons with the energy of [itex]3.68 \cdot 10^{-19}[/itex] J and the wavelength 540 nm. What is the least voltage which must be applied to the diode for it to emit those photons?

Staff: Mentor

Problem: An LED emits photons with the energy of [itex]3.68 \cdot {-19}[/itex] J and the wavelength 540 nm. What is the least voltage which must be applied to the diode for it to emit those photons?

My question: How to find voltage based on this information?

(The answer shall be 2.30 V)

Consider the process that creates the photon: a charge (how big?) falls through an electric potential difference. How much energy does it pick up in doing so? If it is that energy that is converted to light, you should have a relationship to exploit...

The charge comes from recombination of electrons and holes in the diode. Each of these have a charge of 6.242×10^{18} e= 1.602×10^{−19} C and -6.242 ×10^{18} e = -1.602×10^{−19} C, respectively. But how can I know how many does recombine for the diode to emit a photon? From the energy of the emitted photon, which is equal to the energy pisked up when the charge falls through the electric potential difference, right?

Staff: Mentor

One electron --> one photon. You can assume that the hole is stationary (it's "attached" to an atom that's pinned in place in its crystal lattice), so that, essentially, only the electron is accelerating and picking up KE). After all, holes "move" when an electron jumps into it leaving a new vacancy where the electron came from.

Staff: Mentor

In which case, marks awarded might hinge on the quality of the design. So I'd say you consider providing 3 push-button switches: labelled RED, ORANGE, YELLOW

If you want to try something more ambitious, think about powering them with low voltage AC. This way you need only two wires out of the capsule. You supply half-wave DC (of one polarity or the other) to power a single diode, or full-wave to "simultaneously" power both. Employ silicon diodes inside the capsule to do the routing to each led.

In which case, marks awarded might hinge on the quality of the design. So I'd say you consider providing 3 push-button switches: labelled RED, ORANGE, YELLOW

If you want to try something more ambitious, think about powering them with low voltage AC. This way you need only two wires out of the capsule. You supply half-wave DC (of one polarity or the other) to power a single diode, or full-wave to "simultaneously" power both. Employ silicon diodes inside the capsule to do the routing to each led.

Staff: Mentor

Musing further, if diodes were soldered together in parallel (but one reversed) there would be no need for protection diodes, the forward voltage of one would limit the reverse voltage on the other. Needs just 2 LEDs and one resistor. Best to verify this from their data sheets, though, to make sure there is a margin for safety, as significant reverse current though a LED is likely to damage it.

I think I liked this one best... Practically, it would not be difficult to set up. And the plan still will show that one have understood much of the properties of diodes. (I see why it is important to check for a safety margin, so that the reverse current does not damage the LED).

Staff: Mentor

To explain margin of safety: suppose a bright yellow LEDs had a forward drop of 3.8 volts, and your cheap red LED has a reverse voltage rating of 3.6 volts, then the yellow LED would not protect the red one from exceeding its reverse voltage rating.

(I just invented those figures for the purpose of discussion, that forward drop is not typical of current visible-light LEDS.)

To explain margin of safety: suppose a bright yellow LEDs had a forward drop of 3.8 volts, and your cheap red LED has a reverse voltage rating of 3.6 volts, then the yellow LED would not protect the red one from exceeding its reverse voltage rating.

(I just invented those figures for the purpose of discussion, that forward drop is not typical of current visible-light LEDS.)

What the margin of safety would imply is that to protect the red LED from exceeding its reverse voltage rating, the forward drop of the LED has to be a little lower than 3.6 V e.g. 3.4 V or something (in this case), or am I mistaken here?

If that was the case, what about the other way, is it possible that still the red LED will protect the yellow one from exceeding its reverse voltage rating ???

Staff: Mentor

Based on the analysis of your photon energy, I'd say you'll find that the shorter the wavelength coloured LEDs will have a higher forward voltage. So red LEDs have the lowest forward drop of all the visible colours, so would protect all colours. (I'd need to check the data sheets for the different colours, before going ahead and depending on this generalisation, though.)